hmwk4sol_a

hmwk4sol_a - Math 202a Michael Phillips 9/26/07 Problem 1....

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Unformatted text preview: Math 202a Michael Phillips 9/26/07 Problem 1. (a.) Pick any x ∈ [0 , 2 π ]. sin kx has period 2 π k , for any δ > 0, we can take k large enough so that 2 π k < δ- i.e. so that at least two periods of sin kx wholly fit within B ( x,δ ). So, we would then have sin kx = sin k ( x ± 2 π k ) with 2 π k < δ . But then we could choose an x 1 ∈ [ x,x + 2 π k ] such that | sin kx- sin kx 1 | ≥ 1. Hence, we equicontinuity fails since we cannot guarantee the conditions for < 1. (b.) Consider the fixed point x = 1 of all the { x n } . For 0 < x < 1, we know that lim n →∞ = 0. Hence, for any δ > 0, we can find an N so that if n > N , we have that (1- δ ) n < for any > 0. So, we can find an N so that if n > N , we have that | (1) n- (1- δ ) n | is within of one. Hence the conditions for equicontinuity fail at x = 1 if we ask for any < 1. We use Problem 3. We have that the derivatives of the family, i.e. nx n- 1 , are uniformly bounded by 1 in [0 , 1 / 2]. Why? They are all bounded above by 1, the derivative of the constant function at x = 1 / 2. Hence, since each derivative monotonically decreases towards zero, we have that they all stay below 1. Problem 2. Take any 1 > 0. A ( f )( x 1 )- A ( f )( x 2 ) = φ ( x 1 )- φ ( x 2 )+ R b a ( K ( x 1 ,y )- K ( x 2 ,y )) f ( y ) dy . Now since φ is a continuous function on compact domain, we know that it is uniformly continuous on that domain. Likewise, for fixed y , we have that K ( x,y ) is a continuous function of x on a compact domain, hence, is uniformly continuous. So, we have ∀ > 0, a pair δ,δ 1 > 0 so that ∀ x 1 ,x 2 ∈ [ a,b ] we have | ( x 1 ,y )- ( x 2 ,y ) | < δ ⇒ | φ ( x 1 )- φ ( x 2 ) | < and so that | x 1- x 2 | < δ 1 ⇒ | K ( x 1 ,y )- K ( x 2 ,y ) | < . Choose the minimum of these two, call it δ . Now, | x 1- x 2 | < δ , we have | A ( f )( x 1 )- A ( f )( x 2 ) | ≤ + M | a- b | = (1 + M | a- b | ), where M is the bound over all functions in our set of functions in C ([ a,b ]) that is guaranteed by assumption. We want this number to be less than our arbitrary 1 , so we choose δ,δ 1 so that < 1 M | a- b | +1 , then, taking the minimum of δ,δ 1 gives us what we need. Problem 3. Take any > 0. Let δ = /M , where M is such that ∀ f ∈ E and ∀ x ∈ [ a,b ] we have that | f ( x ) | ≤ M . So, suppose | x 1- x 2 | < δ for any x 1 ,x 2 ∈ [ a,b ]. Then | f ( x 1 )- f ( x 2 ) | = | R x 2 x 1 f ( y ) dy | ≤ R x 2 x 1 | f ( y ) | dy ≤ M | x 1- x 2 | ≤ Mδ < , as needed....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

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hmwk4sol_a - Math 202a Michael Phillips 9/26/07 Problem 1....

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